8.6
Sectors and Segments of Circles
In this section we take up issues concerned with portions of circles. We start with two definitions.
Definitions: | A sector
of $\odot O$ is the region bounded by two radii $\overline{OA}$ and
$\overline{OB}$ and one of the arcs $\overparen{AB}$ they determine A segment of a circle is the region bounded by a chord $\overline{CD}$ and the arc $\overparen{CD}$ that it cuts off. |
Note that
if
$A$ and $B$ are two points of $\odot O$, there are, in fact, two sectors $AB$; one uses the minor
arc determined by the points $A$ and $B$, while the other uses the
major arc. Similarly, there are two segments determined by a
chord $\overline{AB}$. We will refer to both sectors and segments
by the degree measure of the arc involved. Most of the time, we
will work with the minor arc. |
In the case of a sector of a circle, the measure of the arc (or, equivalently, of the central angle involved) determines the fraction of the circle we are interested in. In all our dealings with areas of sectors or lengths of arcs, it is this fraction of the circle that we focus on. For example, a $40^\circ$ sector is $\frac{40}{360}$ of the circle. The area of such a sector is $\frac{40}{360}$ of the area of the circle; the length of the arc is $\frac{40}{360}$ of the circumference of the circle.
Example: |
Determine the area
and the perimeter of a $50^\circ$ sector of a circle of radius 12. |
Solution: |
With a $50^\circ$
sector, we are dealing with $\frac{50}{360}$ of the circle. The
area of the sector is $\frac{50}{360}=\frac{5}{36}$ of the area of the
circle. Since the area of the circle is given by $a=\pi \, r^2$,
the area of this circle is $144\pi$. Hence the area of the sector
is $\frac{5}{36}\cdot 144\pi=20\pi$. The circumference of the circle is given by $c=2\pi\,r$. In this case, the circumference of the circle is $24\pi$. The length of the arc of the sector is $\frac{5}{36}\cdot 24\pi=\frac{10\pi}{3}$. To find the perimeter of the sector, we must add two radii. Thus the perimeter of the sector is $\frac{10\pi}{3}+24$. |
Example: | Two circles, of radii 6 and 2
are externally tangent to each other. A tight belt runs around
them, as pictured. What is the length of the belt? |
Solution: | We start by
determining the
length of one of the common external tangent segments. As we have
done before, we put in the line of centers and the radii to the points
of tangency. Then we drop a perpendicular from the center fo the
smaller circle to the radius of the larger one. This produces a
right triangle with a hypotenuse of 8 and a short leg of 4; it is a
30-60-90 triangle! That means that the other leg has length
$4\sqrt{3}$. Therefore $BC=AD=4\sqrt{3}$. In addition, major arc $\overparen{AB}$ on the large circle has degree measure 240, so it represents $\frac{240}{360}$ or $\frac{2}{3}$ of the circumference of the large circle. |
|
And
minor arc $\overparen{CD}$ on the small circle has degree measure 120,
so its
length is $\frac{120}{360}=\frac{1}{3}$ of the circumference of the
smaller circle. Hence the total length of the belt is \begin{align*} 2\cdot 4\sqrt{3}+\frac{2}{3}(2\pi\cdot 6)+\frac{1}{3}(2\pi\cdot 2)&=8\sqrt{3}+8\pi+\frac{4\pi}{3}\\ &=8\sqrt{3}+\frac{28\pi}{3} \end{align*} |
Example: Solution: |
Determine the area
of a $40^\circ$ segment of a circle of radius 6. The area of the sector is $\frac{40}{360}\cdot \pi \cdot 6^2=4\pi$. To get the area of the segment, we must subtract the area of $\triangle AOB$ from the area of the sector. Drop an altitude to $\overline{OB}$ from $A$. Then $\frac{AD}{OA}=\sin 40^\circ$, so that $AD=6\cdot \sin 40^\circ$. Then $a(\triangle AOB)=\frac{1}{2}\cdot 6\cdot 6\sin 40^\circ=11.57$. Finally, the area of the $40^\circ$ segment is $4\pi-11.57=0.996$. |